lc_circuit

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 — lc_circuit [2015/06/07 11:06] (current)admin created 2015/06/07 11:06 admin created 2015/06/07 11:06 admin created Line 1: Line 1: + ====== LC Oscillator Time Domain Analysis ====== + {{:​lc_circuit.png?​400|}} + + ---- + ===== Nodal Equation ===== + + $i_C+i_L=0$ + + $C\frac{dv_{OUT}}{dt}+\frac{1}{L}\int v_{out}=0$ + + $\frac{d^2v_{OUT}}{dt^2}+\frac{1}{LC}v_{OUT}=0$ + + ---- + ===== Solution using method of homogeneous and particular solutions ===== + + First, notice that the nodal equation is a second order homogeneous equation. + Therefore, we will not need to find a particular solution. + + As usual with problems like these, guess that the solution is $e^{st}$. + + Plugging in $v_{OUT=}e^{st}$ into the nodal equation yields: + + $e^{st}\left ( s^2 + \frac{1}{LC} \right )$ + + Solving the above equation for s gives: + + $s=\pm \sqrt{\frac{-1}{LC}}$ + + $s=\pm j\sqrt{\frac{1}{LC}}$ + + So.. + + $v_{out}=e^{j\sqrt{\frac{1}{LC}}t}$ + + and + + $v_{out}=e^{-j\sqrt{\frac{1}{LC}}t}$ + + There are two properties of homogeneous differential equations that are useful. ​ A constant times the solution is still a solution, and the sum of solutions is still a solution. + + Using those two properties we can write: + + $v_{out}=Ae^{j\sqrt{\frac{1}{LC}}t}+Ae^{-j\sqrt{\frac{1}{LC}}t}$ + + Any good math student will recognize that is just, + + $v_{out}=A'​cos \left [t\sqrt{\frac{1}{LC}} \right ]$ + + At $t=0$, $v_{OUT}=v_{IN}$. + + Finally we get, + + $v_{out}=v_{IN}cos \left [t\sqrt{\frac{1}{LC}} \right ]$ + + ---- + + ===== Example ===== + + Let's try to make a 1khz oscillator. ​ + + Remember that $w=2\pi f$. + + $f=\frac{1}{2 \pi\sqrt{LC}}$ + + $1000=\frac{1}{2 \pi\sqrt{LC}}$ + + There are many answers that work...Here is one combination of L and C that produces a 1khz oscillator. ​ + + {{:​lc_simulation.jpg?​800|}}